Question

The conveyor belt is moving at $4m/s$. The coefficient of static friction between the conveyor belt and the $10kg$ package $B$ is ${\mu }_s=0.2$. Determine the shortest time (in $s$) in which the belt can be stopped so that the package does not slide on the belt.

Answer 1

Answer: 2

$a_{\max }={\mu }_{s}g=0.2\times 10=2m/s^2$
$t_{\min }=\frac{u}{a_{\max }}=\frac{4}{2}=2s$